https://file.aiquickdraw.com/mathbot/file/1728595535624vdj4u5bn.jpg

This problem involves determining the orbital period of a satellite around the Earth using Kepler's Third Law. The equation for the orbital period \( T \) of a satellite is derived from Newton's law of gravitation and can be expressed as:

\[
T = 2\pi \sqrt{\frac{r^3}{GM}}
\]

Where:
- \( T \) is the orbital period,
- \( r \) is the orbital radius (distance from the center of the Earth to the satellite),
- \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \),
- \( M \) is the mass of the Earth.

### Given dataThis problem involves determining the orbital period of a satellite around the Earth using Kepler's Third Law. The equation for the orbital period \( T \) of a satellite is derived from Newton's law of gravitation and can be expressed as:

\[
T = 2\pi \sqrt{\frac{r^3}{GM}}
\]

Where:
- \( T \) is the orbital period,
- \( r \) is the orbital radius (distance from the center of the Earth to the satellite),
- \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \),
- \( M \) is the mass of the Earth.

### Given 
- Distance of the satellite from Earth: \( r = 3.93 \times 10^8 \, \text{m} \),
- Earth's mass: \( M = 5.97 \times 10^{24} \, \text{kg} \),
- Satellite's mass: \( m = 4.68 \times 10^3 \, \text{kg} \) (but this isn't needed since the satellite's mass doesn't affect the orbital period).

### Calculation:

1. Plug the values into the formula:

\[
T = 2\pi \sqrt{\frac{(3.93 \times 10^8)^3}{6.674 \times 10^{-11} \times 5.97 \times 10^{24}}}
\]

2. Compute the expression to find \( T \) (orbital period).

I'll now calculate the orbital period for you.The orbital period of the satellite is approximately \( 2,452,384 \) seconds.

To express this in a more common time format:
- \( 2,452,384 \, \text{seconds} \) is about 28.39 days.

Would you like further details on the calculation or explanation? 

Here are some related questions you might consider:
1. How does the mass of a satellite affect its orbit?
2. What is the significance of Kepler's Third Law in satellite orbits?
3. How would the period change if the satellite were farther from Earth?
4. How do geostationary satellites differ in orbital period?
5. What role does the gravitational constant play in orbital mechanics?

**Tip:** Orbital period depends on the distance from the center of the Earth, not the satellite's mass!

A satellite is about 3.93×10^8 m from the Earth, along the semi-major axis of its orbit. If the satellite's mass is 4.68×10^3 kg, and the Earth's mass is about 5.97×10^24 kg, what is the period of the satellite's orbit around the Earth? Express your answer in s and in decimal form, not scientific notation.

Determine the orbital period of a satellite 3.93×10^8 m from Earth with a mass of 4.68×10^3 kg, using the formula T = 2π√(r^3/GM). Learn how to apply Kepler's Third Law for accurate results, suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation